scala how to reduce while n > 0 - scala

Question

I am wondering if there's a way to deal with a while (n > 0) loop in a more functional way, I have a small Scala app that counts the number of digits equal to K from a range from 1 to N:
for example 30 and 3 would return 4 [3, 13, 23, 30]
object NumKCount {
def main(args: Array[String]): Unit = {
println(countK(30,3))
}
def countKDigit(n:Int, k:Int):Int = {
var num = n
var count = 0
while (num > 10) {
val digit = num % 10
if (digit == k) {count += 1}
num = num / 10
}
if (num == k) {count += 1}
count
}
def countK(n:Int, k:Int):Int = {
1.to(n).foldLeft(0)((acc, x) => acc + countKDigit(x, k))
}
}
I'm looking for a way to define the function countKDigit using a purely functional approach

Answer 1

First expand number n into a sequence of digits
def digits(n: Int): Seq[Int] = {
if (n < 10) Seq(n)
else digits(n / 10) :+ n % 10
}
Then reduce the sequence by counting occurrences of k
def countKDigit(n:Int, k:Int):Int = {
digits(n).count(_ == k)
}
Or you can avoid countKDigit entirely by using flatMap
def countK(n:Int, k:Int):Int = {
1.to(n).flatMap(digits).count(_ == k)
}

Answer 2

Assuming that K is always 1 digit, you can convert n to String and use collect or filter, like below (there's not much functional stuff you can do with Integer):
def countKDigit(n: Int, k: Int): Int = {
n.toString.collect({ case c if c.asDigit == k => true }).size
}
or
def countKDigit(n: Int, k: Int): Int = {
n.toString.filter(c => c.asDigit == 3).length
}
E.g.
scala> 343.toString.collect({ case c if c.asDigit == 3 => true }).size
res18: Int = 2
scala> 343.toString.filter(c => c.asDigit == 3).length
res22: Int = 2

Answer 3

What about the following approach:
scala> val myInt = 346763
myInt: Int = 346763
scala> val target = 3
target: Int = 3
scala> val temp = List.tabulate(math.log10(myInt).toInt + 1)(x => math.pow(10, x).toInt)
temp: List[Int] = List(1, 10, 100, 1000, 10000, 100000)
scala> temp.map(x => myInt / x % 10)
res17: List[Int] = List(3, 6, 7, 6, 4, 3)
scala> temp.count(x => myInt / x % 10 == target)
res18: Int = 2

Answer 4

Counting the occurrences of a single digit in a number sequence.
def countK(n:Int, k:Int):Int = {
assert(k >= 0 && k <= 9)
1.to(n).mkString.count(_ == '0' + k)
}
If you really only want to modify countKDigit() to a more functional design, there's always recursion.
def countKDigit(n:Int, k:Int, acc: Int = 0):Int =
if (n == 0) acc
else countKDigit(n/10, k, if (n%10 == k) acc+1 else acc)